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IAL 2020 Oct Q8

A Level / Edexcel / P3

IAL 2020 Oct Paper · Question 8

题目

Problem

(i) The curve CC has equation y=g(x)y=g(x) where

g(x)=e3xsec⁡2x−π4<x<π4g(x)=e^{3x}\sec 2x\qquad -\frac{\pi}{4}<x<\frac{\pi}{4}

(a) Find g′(x)g'(x).

(2)

(b) Hence find the xx coordinate of the stationary point of CC.

(3)

(ii) A different curve has equation

x=ln⁡(sin⁡y)0<y<π2x=\ln(\sin y)\qquad 0<y<\frac{\pi}{2}

Show that

dydx=exf(x)\frac{dy}{dx}=\frac{e^x}{f(x)}

where f(x)f(x) is a function of exe^x that should be found.

(4)
题目中文翻译

(i) 曲线 CC 的方程为 y=g(x)y=g(x)，其中

g(x)=e3xsec⁡2x−π4<x<π4g(x)=e^{3x}\sec 2x\qquad -\frac{\pi}{4}<x<\frac{\pi}{4}

(a) 求 g′(x)g'(x)。

(b) 由此求曲线 CC 的驻点的 xx 坐标。

(ii) 另一条曲线的方程为

x=ln⁡(sin⁡y)0<y<π2x=\ln(\sin y)\qquad 0<y<\frac{\pi}{2}

证明

dydx=exf(x)\frac{dy}{dx}=\frac{e^x}{f(x)}

其中 f(x)f(x) 是 exe^x 的某个函数，并求出该函数。

解答

(i)(a)

We have

g(x)=e3xsec⁡2xg(x)=e^{3x}\sec2x

Use the product rule:

g′(x)=ddx(e3x)sec⁡2x+e3xddx(sec⁡2x)g'(x)=\frac{\mathrm{d}}{\mathrm{d}x}(e^{3x})\sec2x+e^{3x}\frac{\mathrm{d}}{\mathrm{d}x}(\sec2x)

Now

ddx(e3x)=3e3x\frac{\mathrm{d}}{\mathrm{d}x}(e^{3x})=3e^{3x}

and

ddx(sec⁡2x)=2sec⁡2xtan⁡2x\frac{\mathrm{d}}{\mathrm{d}x}(\sec2x)=2\sec2x\tan2x

Therefore

g′(x)=3e3xsec⁡2x+2e3xsec⁡2xtan⁡2x\boxed{g'(x)=3e^{3x}\sec2x+2e^{3x}\sec2x\tan2x}

This can also be written as

g′(x)=e3xsec⁡2x(3+2tan⁡2x)\boxed{g'(x)=e^{3x}\sec2x(3+2\tan2x)}

(i)(b)

At a stationary point,

g′(x)=0g'(x)=0

Using the factorised form,

e3xsec⁡2x(3+2tan⁡2x)=0e^{3x}\sec2x(3+2\tan2x)=0

For

−π4<x<π4-\frac{\pi}{4}<x<\frac{\pi}{4}

we have

e3x≠0,sec⁡2x≠0e^{3x}\ne 0,\qquad \sec2x\ne 0

So

3+2tan⁡2x=03+2\tan2x=0

Therefore

tan⁡2x=−32\tan2x=-\frac32

So

2x=arctan⁡(−32)2x=\arctan\left(-\frac32\right)

and hence

x=12arctan⁡(−32)x=\frac12\arctan\left(-\frac32\right)

Using a calculator,

x=−0.491396…x=-0.491396\ldots

Therefore

x=−0.491\boxed{x=-0.491}

to 3 significant figures.

(ii)

We are given

x=ln⁡(sin⁡y)x=\ln(\sin y)

Exponentiate both sides:

ex=sin⁡ye^x=\sin y

Since

0<y<π20<y<\frac{\pi}{2}

we have cos⁡y>0\cos y>0. Therefore

cos⁡y=1−sin⁡2y\cos y=\sqrt{1-\sin^2y}

Using sin⁡y=ex\sin y=e^x,

cos⁡y=1−e2x\cos y=\sqrt{1-e^{2x}}

Now differentiate

x=ln⁡(sin⁡y)x=\ln(\sin y)

with respect to yy:

dxdy=cos⁡ysin⁡y\frac{\mathrm{d}x}{\mathrm{d}y}=\frac{\cos y}{\sin y}

So

dydx=sin⁡ycos⁡y\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\sin y}{\cos y}

Substitute

sin⁡y=ex,cos⁡y=1−e2x\sin y=e^x,\qquad \cos y=\sqrt{1-e^{2x}}

to get

dydx=ex1−e2x\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{e^x}{\sqrt{1-e^{2x}}}

Therefore

dydx=exf(x)\boxed{\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{e^x}{f(x)}}

where

f(x)=1−e2x\boxed{f(x)=\sqrt{1-e^{2x}}}